Question

A cell with N/50 KCL solutions showed a resistance of 550 ohms at 25 degree c The specific conductivity of N/50 KCL at 25 degree c is 0.002768 ohm's cm the cell filled with N/50 Znso4 solution at 25 degreec shows resistance of 72.18 ohms find the cell constant and molar conductivity of Znso4 solution

Answer 1

Given in the question.
Resistance = 550 ohm.
Specific conductivity = 0.002768 ohm⁻¹ cm⁻¹

First we have to find the value of Cell constant . 
Cell constant =specific conductivity × resistance {K = Cell const./Resistance}
Cell constant = 0.002768 × 550
= 1.5224 .

For ZnSO₄ , Conductance = 1/R 
=1/72.18
= 0.014

Specific conductance = cell constant × resistance 
= 1.5224 × 0.014
= 0.021 .


Now finding the molar conductance.
Λ = κ × 1000 / M. --------eq(i). 

Where M is the molarity of ZnSO₄  
Now, 
∵ Molarity = Normality × n factor = (1/10) × 2
∴ Molarity = 0.2 M

Putting the value of the molarity in eq(i). 

Λ = 0.021 × 1000 / 0.2 
  = 21/0.2
  = 105 

Hence, molar conductivity of the solution is 105 seimens/m. 



Hope it helps.